E15

Moderators: Chem_Mod, Chem_Admin

ChristianM3F
Posts: 61
Joined: Sat Sep 14, 2019 12:16 am

E15

Postby ChristianM3F » Wed Oct 02, 2019 4:29 pm

Quick question about E15... Am I missing something here? They ask what the molar mass of the sulfide in M(OH)2 is, but doesn't a sulfide need a sulfur molecule? (Chemistry's hard...

Simon Dionson 4I
Posts: 107
Joined: Sat Sep 14, 2019 12:17 am

Re: E15

Postby Simon Dionson 4I » Wed Oct 02, 2019 4:40 pm

I agree I'm pretty sure the metal in question isn't a sulfide because its molar mass is greater. Maybe there was a typo or something we missed in the problem?

karinaseth_1A
Posts: 57
Joined: Sat Jul 20, 2019 12:15 am

Re: E15

Postby karinaseth_1A » Wed Oct 02, 2019 4:42 pm

When they ask for the molar mass of the sulfide, they mean that you have to swap out the hydroxide - the OH - for a sulfur atom. Because you're given the molar mass of the metal hydroxide, you can figure out the molar mass of the metal 'M' by subtracting the molar mass of two hydroxide molecules (17.01 g/mol x 2 = 34.02 g/mol) from the given molar mass which is 74.10 g/mol. Doing this, you find out that the molar mass of the metal 'M' is 40.08 g/mol. Then, you can add the molar mass of sulfur to that (32.07 g/mol) and find out that the sulfide of the metal 'M' (MS) has a molar mass of 72.15 g/mol.

AArmellini_1I
Posts: 107
Joined: Fri Aug 09, 2019 12:15 am

Re: E15

Postby AArmellini_1I » Wed Oct 02, 2019 4:45 pm

I was confused with that question too. This is how I processed it. M = the metal = Calcium (because 74.10 g/mol - 34.02 g/mol = 40.08 g/mol, which is calcium's atomic mass. The questions asks "What is the molar mass of the sulfide of this metal?", so the book is no longer referencing the equation but asking you what the molar mass of calcium + molar mass of sulfur is (CaS). So the sulfide of our metal would have the molar mass is 72.15g/mol (40.08 g/mol + 32.065 g/mol). I hope that makes sense. Sulfur is not suppose to be represented in the given chemical equation.


Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 2 guests